Math Problem Statement

Old Time Savings Bank pays 5.75% interest on its savings account. If you deposit $1,000 in the bank and leave it there:

How much interest will you earn in the first year?

Note: Round your answer to 2 decimal places.

How much interest will you earn in the second year?

Note: Do not round intermediate calculations. Round your answer to 2 decimal places.

How much interest will you earn in the tenth year?

Note: Do not round intermediate calculations. Round your answer to 2 decimal places.

Solution

To solve this, we’ll assume that the interest is compounded annually, meaning that after each year, the interest is added to the principal amount, and the next year’s interest is calculated on the new total.

The formula for compound interest is:

A=P(1+rn)ntA = P \left(1 + \frac{r}{n}\right)^{nt}

Where:

  • AA = the amount of money after interest
  • PP = the principal amount (initial deposit)
  • rr = the annual interest rate (expressed as a decimal)
  • nn = the number of times the interest is compounded per year
  • tt = the number of years the money is invested

In this case:

  • P=1000P = 1000
  • r=5.75%r = 5.75\% or 0.0575
  • n=1n = 1 (since interest is compounded annually)
  • t=1t = 1 for the first year, t=2t = 2 for the second year, and t=10t = 10 for the tenth year.

Interest earned in the first year:

For t=1t = 1, A1=1000(1+0.05751)1×1=1000×(1.0575)=1057.50A_1 = 1000 \left(1 + \frac{0.0575}{1}\right)^{1 \times 1} = 1000 \times (1.0575) = 1057.50 Interest earned in the first year: Interest=A1P=1057.501000=57.50\text{Interest} = A_1 - P = 1057.50 - 1000 = 57.50

Interest earned in the second year:

For t=2t = 2, A2=1000(1+0.05751)1×2=1000×(1.0575)2=1118.80A_2 = 1000 \left(1 + \frac{0.0575}{1}\right)^{1 \times 2} = 1000 \times (1.0575)^2 = 1118.80 Interest earned in the second year: Interest=A2A1=1118.801057.50=61.30\text{Interest} = A_2 - A_1 = 1118.80 - 1057.50 = 61.30

Interest earned in the tenth year:

For t=10t = 10, A10=1000(1+0.05751)1×10=1000×(1.0575)10=1769.31A_{10} = 1000 \left(1 + \frac{0.0575}{1}\right)^{1 \times 10} = 1000 \times (1.0575)^{10} = 1769.31 Interest earned in the tenth year: Interest=A10A9=1769.311674.04=95.27\text{Interest} = A_{10} - A_9 = 1769.31 - 1674.04 = 95.27

Summary:

  1. Interest earned in the first year: $57.50
  2. Interest earned in the second year: $61.30
  3. Interest earned in the tenth year: $95.27

Would you like more details or explanations about these calculations?


Here are 5 related questions for further exploration:

  1. What will be the total amount in your account after 5 years with the same interest rate?
  2. How would the interest change if the bank offered monthly compounding instead of annual?
  3. How much would you earn in the second year if you deposited $2,000 instead of $1,000?
  4. What would the interest be after 20 years with the same rate and compounding frequency?
  5. How would a decrease in the interest rate to 4.5% affect your total earnings after 10 years?

Tip: Compound interest grows faster with more frequent compounding periods (e.g., monthly instead of annually).

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Math Problem Analysis

Mathematical Concepts

Compound Interest
Exponential Growth

Formulas

A = P(1 + r/n)^(nt)

Theorems

Compound Interest Theorem

Suitable Grade Level

Grades 9-12